Solve-5-log-x-logx-5-25-

Question Number 13152 by tawa tawa last updated on 15/May/17

Solve : 5^{\log (x)} + {logx}^{5} = 25

Answered by mrW1 last updated on 16/May/17

\log x = t

5^{t} + 5 t = 25

5^{t} = 5 (5 - t)

5^{t - 5} = \frac{5 - t}{625}

e^{(t - 5) \ln 5} = \frac{5 - t}{625}

(5 - t) \ln 5 e^{(t - 5) \ln 5} = 625 \times \ln 5

(5 - t) \ln 5 = W (625 \times \ln 5)

t = 5 - \frac{W (625 \times \ln 5)}{\ln 5} = \log x

\Rightarrow x = 10^{5 - \frac{W (625 \times \ln 5)}{\ln 5}} \approx 10^{5 - \frac{5.254544}{\ln 5}} \approx 54.346

Commented by tawa tawa last updated on 15/May/17

Wow, God bless you sir .